Chapter 38, Problem 37P

(a)

To determine

The value of mean free path ${\lambda }_{\text{i}}$ .

Answer to Problem 37P

The value of mean free path ${\lambda }_{\text{i}}$ is $66\text{nm}$ .

Explanation of Solution

Given:

The resistivity is $\rho =1.0\times {10}^{-8}\text{Ω}\cdot \text{m}$ .

The Fermi energy from Table 38-1 is $E_F=7.03\text{eV}$ .

The value of $n_e\left(=N/V\right)$ for copper from Table 38-1 is $n_e=84.7\text{electrons}/{\text{nm}}^{\text{3}}$ .

Formula used:

The expression of resistivity from equation 38-13 is,

  $\begin{array}{c}\rho =\frac{m_e{v}_{av}}{n_e{e}^2\lambda }\\ =\frac{m_e{u}_F}{n_e{e}^2{\lambda }_{\text{i}}}\end{array}$   ...... (1)

The expression for Fermi speed is,

  $u_F=\sqrt{\frac{2E_F}{m_e}}$   ...... (2)

Calculation:

From equation (1), the mean free path is,

  $\begin{array}{c}{\lambda }_{\text{i}}=\frac{m_e{u}_F}{n_e{e}^2\rho }\\ =\frac{\sqrt{2m_e{E}_F}}{n_e{e}^2\rho }\end{array}$

Substitute values in above expression,

  $\begin{array}{c}{\lambda }_{\text{i}}=\frac{\sqrt{2\left(9.109\times {10}^{-31}\text{kg}\right)\left(7.03\text{eV}\right)}}{\left(84.7\text{electrons}/{\text{nm}}^{\text{3}}\right){\left(1.602\times {10}^{-19}\text{C}\right)}^2\left(1.0\times {10}^{-8}\text{Ω}\cdot \text{m}\right)}\\ =\frac{\sqrt{2\left(9.109\times {10}^{-31}\text{kg}\right)\left(7.03\text{eV}\right)\left(1.602\times {10}^{-19}\text{J}/\text{eV}\right)}}{\left(84.7\times {10}^{27}\text{electrons}/{\text{m}}^{\text{3}}\right){\left(1.602\times {10}^{-19}\text{C}\right)}^2\left(1.0\times {10}^{-8}\text{Ω}\cdot \text{m}\right)}\\ =66\text{nm}\end{array}$

Conclusion:

Therefore, the value of mean free path ${\lambda }_{\text{i}}$ is $66\text{nm}$ .

(b)

To determine

The value of scattering cross section area.

Answer to Problem 37P

The value of scattering cross section area is $1.8\times {10}^{-4}{\text{nm}}^{\text{2}}$ .

Explanation of Solution

Formula used:

The expression for mean free path from equation 38-16 is,

  ${\lambda }_{\text{i}}=\frac{1}{n_{e}A}$   ...... (3)

Here, $A$ is the scattering cross section area.

From equation (3), the scattering cross section area is,

  $A=\frac{1}{n_e{\lambda }_{\text{i}}}$

Substitute values in above equation,

  $\begin{array}{c}A=\frac{1}{\left(84.7\text{electrons}/{\text{nm}}^{\text{3}}\right)\left(66\text{nm}\right)}\\ =1.788\times {10}^{-4}{\text{nm}}^{\text{2}}\\ \approx 1.8\times {10}^{-4}{\text{nm}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the value of scattering cross section area is $1.8\times {10}^{-4}{\text{nm}}^{\text{2}}$ .